Let A be a countable set and B be an uncountable set. (i) Is

Let A be a countable set and B be an uncountable set. (i) Is
countable or uncountable? Explain your answer. (ii) Is
countable or uncountable? Explain your answer. (iii) Is
countable or uncountable? Explain your answer. [15 marks]

The Correct Answer and Explanation is:

Here is the solution presented in a formal, textbook-style format.

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Let A be a countable set and B be an uncountable set.

(i) Is 

A∪BA∪B

 countable or uncountable?

The union 

A∪BA∪B

 is uncountable.

Explanation:
The union of two sets, 

A∪BA∪B

, contains all the elements of both A and B. By definition, the set B is a subset of the union 

A∪BA∪B

, which can be written as 

B⊆A∪BB⊆A∪B

.

A fundamental theorem in set theory states that any set which contains an uncountable subset must itself be uncountable. If 

A∪BA∪B

 were countable, then every subset of it would also have to be countable. This would imply that B is countable, which contradicts the initial premise that B is an uncountable set. Therefore, the assumption that 

A∪BA∪B

 is countable must be false. It follows that the union 

A∪BA∪B

 is uncountable. In essence, adding the elements of a countable set to an uncountable set does not change the uncountability of the resulting set.

(ii) Is 

A∩BA∩B

 countable or uncountable?

The intersection 

A∩BA∩B

 is countable.

Explanation:
The intersection of two sets, 

A∩BA∩B

, consists of all elements that are common to both A and B. By definition, every element of the intersection must be an element of A. This means that the intersection 

A∩BA∩B

 is a subset of A, written as 

A∩B⊆AA∩B⊆A

.

A key property of countable sets is that any subset of a countable set is also countable. Since A is given as a countable set, any of its subsets, including 

A∩BA∩B

, must be countable. This holds true regardless of the nature of B. The intersection could be empty, finite, or countably infinite, but in all possible cases, it cannot be uncountable because it is bounded by the size of the countable set A.

(iii) Is 

A×BA×B

 countable or uncountable?

The Cartesian product 

A×BA×B

 is uncountable.

Explanation:
This conclusion assumes A is non-empty. (If A were the empty set, 

A×BA×B

 would also be empty and thus countable). Since A is a non-empty countable set, it contains at least one element. Let 

a0a0​

 be a fixed, arbitrary element in A.

Consider the subset of the Cartesian product defined as 

S={a0}×BS={a0​}×B

. This set consists of all ordered pairs of the form 

(a0,b)(a0​,b)

, where 

bb

 is any element of B. A one-to-one correspondence (a bijection) can be established between the set B and the set S by the mapping 

f(b)=(a0,b)f(b)=(a0​,b)

. Since there is a bijection between B and S, and B is uncountable, the set S must also be uncountable.

The set S is a subset of the full Cartesian product 

A×BA×B

 (i.e., 

S⊆A×BS⊆A×B

). As established in part (i), any set that has an uncountable subset is itself uncountable. Since 

A×BA×B

 contains the uncountable subset S, the Cartesian product 

A×BA×B

 must be uncountable.